Problem: $f(x, y, z) = \cos(x + y) - \sin(z - y)$ What is the Laplacian of $f(x, y, z)$ ? $\Delta f = $
Explanation: The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\Delta f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}$ [What does that triangle mean?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ -\sin(x + y) \right] \\ \\ &= -\cos(x + y) \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ -\sin(x + y) + \cos(z - y) \right] \\ \\ &= -\cos(x + y) + \sin(z - y) \\ \\ f_{zz} &= \dfrac{\partial}{\partial z} \left[ \dfrac{\partial f}{\partial z} \right] \\ \\ &= \dfrac{\partial}{\partial z} \left[ -\cos(z - y) \right] \\ \\ &= \sin(z - y) \end{aligned}$ The Laplacian is $\Delta f = f_{xx} + f_{yy} + f_{zz}$. Therefore: $\Delta f(x, y, z) = -2\cos(x + y) + 2\sin(z - y)$